Problem: Takahashi loves sorting.

He has a permutation (p_1,p_2,...,p_N) of the integers from 1 through N. Now, he will repeat the following operation until the permutation becomes (1,2,...,N):

* First, we will define high and low elements in the permutation, as follows. The i-th element in the permutation is high if the maximum element between the 1-st and i-th elements, inclusive, is the i-th element itself, and otherwise the i-th element is low.
* Then, let a_1,a_2,...,a_k be the values of the high elements, and b_1,b_2,...,b_{N-k} be the values of the low elements in the current permutation, in the order they appear in it.
* Lastly, rearrange the permutation into (b_1,b_2,...,b_{N-k},a_1,a_2,...,a_k).



How many operations are necessary until the permutation is sorted?

Constraints

* 1 ≤ N ≤ 2×10^5
* (p_1,p_2,...,p_N) is a permutation of the integers from 1 through N.

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 ... p_N


Output

Print the number of operations that are necessary until the permutation is sorted.

Examples

Input

5
3 5 1 2 4


Output

3


Input

5
5 4 3 2 1


Output

4


Input

10
2 10 5 7 3 6 4 9 8 1


Output

6

Solution (in CPP):

Solution:
#include <cstdio>
#include <algorithm>

using namespace std;

namespace io
{
	const int maxb = 1 << 15;
	char b[maxb], *s = b, *t = b;
	bool Getchar(char &ch)
	{
		return ch = s == t && (t = (s = b) + fread(b, 1, maxb, stdin)) == b ? 0 : *s ++;
	}
	int Getint()
	{
		char ch;
		while (Getchar(ch) && (ch < '0' || ch > '9'));
		int s = ch - '0';
		while (Getchar(ch) && ch >= '0' && ch <= '9')
			s = s * 10 + ch - '0';
		return s;
	}
}

using io::Getint;

const int maxn = 2e5 + 10;

int p[maxn], q[maxn];

bool Check(int a, int b, int c)
{
	if (a < b && b < c) return true;
	if (b < c && c < a) return true;
	if (c < a && a < b) return true;
	return false;
}

int main()
{
	int n = Getint();
	for (int i = 1; i <= n; ++ i)
		q[p[i] = Getint()] = i;
	int T = 0, f;
	for (int i = n - 1; i; -- i)
	{
		if (!T)
		{
			if (q[i] < q[i + 1]) T = 0;
			else ++ T, f = i + 1;
		}
		else
			if (!Check(q[f], q[i], q[i + 1])) ++ T, f = i + 1;
	}
	printf("%d\n", T);
	return 0;
}